Expert Answer to "If a and are any odd integers, then..."

djeljenike

Answered question

2020-10-27

If a and are any odd integers, then

a^{2} + b^{2}

is even.
Prove?

Answer & Explanation

l1koV

Skilled2020-10-28Added 100 answers

Proof:
Let two odd positive numbers be $a = 2 j + 1 a n d b = 2 k + 1 :$ where j and k are integers.
$a^{2} + b^{2} = (2 j + 1) 2 + (2 k + 1)^{2}$
$= 4 j^{2} + 4 j + 1 + 4 k^{2} + 4 k + 1$
$= 2 (2 j^{2} + 2 j) + 1 + 2 (2 k^{2} + 2 k) + 1$
$= 2 l + 2 m + 2$
where, $l = 2 j^{2} + 2 j, m = 2 k^{2} + 2 k$ are integers. Then,
$a^{2} + b^{2} = 2 (l + m + 1)$
$a^{2} + b^{2} = 2 n$ which is even by definition.
where, $n = l + m + 1$ is also an integer.
Hence, if a and b are any odd integers then $a^{2} + b^{2}$ is even.
Conclusion:
The statement, "If a and b are any odd integers, then $a^{2} + b^{2}$ is even" is proved.

Jeffrey Jordon

Expert2021-11-03Added 2605 answers

Answer is given below (on video)

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